Solve y=8x/5-tan(x) | Microsoft Math Solver

Solve for y (complex solution)

y=\frac{16ix\cos(x)}{\left(-1+5i\right)e^{ix}+\left(1+5i\right)e^{-ix}}

\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}+\arctan(5)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}

Solve for y

y=\frac{8x\cos(x)}{5\cos(x)-\sin(x)}

\nexists n_{3}\in \mathrm{Z}\text{ : }x=\pi n_{3}+\arcsin(\frac{5\sqrt{26}}{26})\text{ and }\nexists n_{2}\in \mathrm{Z}\text{ : }x=\pi n_{2}+\arctan(5)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}

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Trigonometry

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y = \frac { 8 x } { 5 - \tan ( x ) }

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How do you differentiate \displaystyle{y}=\frac{{x}}{{{2}-{\tan{{x}}}}} ?

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\displaystyle\frac{{{2}-{\tan{{x}}}+{x}{{\sec}^{{2}}{x}}}}{{\left({2}-{\tan{{x}}}\right)}^{{2}}} Explanation: \displaystyle{y}=\frac{{x}}{{{2}-{\tan{{x}}}}} \displaystyle{\left(\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(\frac{{u}}{{v}}\right)}=\frac{{{v}\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}-{u}\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}}{{v}^{{2}}}\right)} ...

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Domain: \displaystyle{x}\ne\frac{{k}}{{2}}\pi,{k}={0},\pm{1},\pm{2},\pm{3},.. Range: \displaystyle{\left(-\infty,\infty\right)} . Asymptotes: \displaystyle{x}=\frac{{k}}{{2}}\pi,{k}={0},\pm{1},\pm{2},\pm{3},.. ...

What is the implicit derivative of \displaystyle{1}=\frac{{x}}{{y}}-{x}{\tan{{y}}} ?

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\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}=-\frac{{{y}{\left({y}{\tan{{y}}}-{1}\right)}}}{{{x}{\left({1}+{y}^{{2}}{{\sec}^{{2}}{y}}\right)}}} Explanation: \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{\left[{1}=\frac{{x}}{{y}}-{x}{\tan{{y}}}\right]} ...

How do you find the amplitude and period of a function \displaystyle{y}=-{\tan{{\left({x}-\frac{\pi}{{2}}\right)}}} ?

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An explanation is given below. Explanation: The \displaystyle{A}{\tan{{\left({B}{\left({x}-{C}\right)}\right)}}}+{D} function has \displaystyle\text{no amplitude.} The period is given by \displaystyle\frac{\pi}{{B}} ...

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If you multiply your numerator and denominator by \cos^2 x and use \sin^2 x=1-\cos^2 x, the expression becomes \frac {3-4\cos^2 x}{1-4\cos^2 x} Now set z=4\cos^2 x and note that 0\le z \le 4 ...

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\displaystyle-\frac{{{y}^{{2}}-{y}{{\sec}^{{2}}{\left(\frac{{x}}{{y}}\right)}}}}{{{x}{{\sec}^{{2}}{\left(\frac{{x}}{{y}}\right)}}+{y}^{{2}}}} Explanation: \displaystyle\text{differentiate }\ {\tan{{\left(\frac{{x}}{{y}}\right)}}}\ \text{ using the }\ \text{chain rule} ...

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